圆的方程为 $x^2 + y^2 + 8x - 20y - 53 = 0$。

1. Group the $x$ and $y$ terms and complete the square for both.

$x^2 + 8x + y^2 - 20y = 53$
$(x + 4)^2 - 4^2 + (y - 10)^2 - (-10)^2 = 53$
$(x + 4)^2 - 16 + (y - 10)^2 - 100 = 53$
$(x + 4)^2 + (y - 10)^2 = 169$

2. Identify the coordinates of the centre from the standard form $(x-a)^2 + (y-b)^2 = r^2$.

Centre is $(-4, 10)$

3. Calculate the radius from the standard form.

$r^2 = 169$
$r = \sqrt{169} = 13$

1. State the coordinates of the centre of the circle from part (i)

From the equation $x^2 + y^2 + 8x - 20y - 53 = 0$, the centre $C$ is $(-4, 10)$.

2. Calculate the gradient of the normal (the radius connecting the centre to $P$)

Gradient of $CP = \frac{-2 - 10}{1 - (-4)} = \frac{-12}{5}$

3. Determine the gradient of the tangent

Since the tangent is perpendicular to the radius at the point of contact,
Gradient of tangent $m = -\frac{1}{\left(\frac{-12}{5}\right)} = \frac{5}{12}$

4. Find the equation of the tangent line

Equation of the tangent at $P(1, -2)$:
$y - (-2) = \frac{5}{12}(x - 1)$
$12(y + 2) = 5(x - 1)$
$12y + 24 = 5x - 5$
$5x - 12y - 29 = 0$

1. Find the gradient of the radius connecting the centre to $Q$

From (i), the centre of the circle is $C(-4, 10)$.
Gradient of radius $CQ = \frac{10 - 10}{9 - (-4)} = 0$.

2. Explain the relationship between the radius and the tangent

Since the gradient of the radius $CQ$ is $0$, the radius is a horizontal line (parallel to the $x$-axis).
The tangent to the circle at $Q$ is perpendicular to the radius at $Q$.
Therefore, the tangent is a vertical line, which means it is parallel to the $y$-axis.